Does a simple pendulum of length 0.50 m have a larger or smaller frequency of vi

Question

Does a simple pendulum of length 0.50 m have a larger or smaller frequency of vibration than a simple pendulum of length 1.0 m? Explain. (Select all that apply.) Smaller. A longer pendulum has a longer period and smaller frequency according to omega = Squareroot mg/(mL) = Squareroot g/L Smaller. A longer string makes the restoring force smaller for each pendulum angle. Larger. It increases the distance the pendulum bob rises as it swings through the same angle, doubling its frequency. Larger. It increases the distance the pendulum bob must move along its path, making it move faster. There is no change. The frequency of a pendulum does not depend on its amplitude. Examine the derivation of the frequency of a pendulum. Consider whether the restoring force F_t is increased or decreased by making the string longer Does the lack of dependence on amplitude imply a lack of dependence on length? Use the worked example above to help you solve this problem. Using a small pendulum of length 0.174 m, a geophysicist counts 70.0 complete swings in a time of 60.0 s. What is the value of g in this location? What would be the period of the same pendulum (from the PRACTICE IT section) on the Moon, where the acceleration of gravity is 1.62 m/s^2? T = Follow the example, but not too closely. Instead of solving for g in terms of T, you need to solve for T in terms of gravity is 1.62 m/s^2? Follow the example, but not too closely. Instead of solving for g in terms of T, you need to solve for T in terms of g.

Explanation / Answer

we know

T = 2pi * sqrt(L/g)

L = 0.174 m

g = 1.62 m/s^2

T = 2*pi * sqrt(0.174/1.62)

T = 2.059 s

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